Evaluate this definite integral

[utkarshakash]

Homework Statement

$$ \displaystyle \int_0^{\infty} e^{-x} \dfrac{a\sin ax - \cos ax}{1+a^2} da $$

Homework Equations

The Attempt at a Solution

Evaluating this using integration by parts will be a cumbersome process and I don't even think that would give me the answer. Substitutions aren't going to make this problem easier. What other methods should I apply?

 

[Simon Bridge]

simplify or change the form of the integrand ... i.e. look for an identity involving the trig functions.
the euler formula perhaps ... what have you tried?

[edit] Ugh - looking up the indefinite integrals ... it's nasty.
See "trigonometric integrals"
http://en.wikipedia.org/wiki/Trigonometric_integral

Where does this come from?

 

[SteamKing]

Is the variable of integration 'a' instead of 'x'?

 

[utkarshakash]

Is the variable of integration 'a' instead of 'x'?

Yes

 

[Curious3141]

Homework Statement

$$ \displaystyle \int_0^{\infty} e^{-x} \dfrac{a\sin ax - \cos ax}{1+a^2} da $$

Homework Equations

The Attempt at a Solution

Evaluating this using integration by parts will be a cumbersome process and I don't even think that would give me the answer. Substitutions aren't going to make this problem easier. What other methods should I apply?

Differentiating under the integral sign might help here, I think.

$$\displaystyle \frac{d}{dx}\int_0^{\infty} e^{-x} \dfrac{a\sin ax - \cos ax}{1+a^2} da = \int_0^{\infty} \frac{\partial}{\partial x} (e^{-x} \dfrac{a\sin ax - \cos ax}{1+a^2}) da = \int_0^{\infty}e^{-x}\cos ax da $$ which is non convergent.

So the original integral also does not exist.

I'm not 100% sure about this, though.

 

[haruspex]

Yes

Then I don't understand the significance of the exponential term. Why is that not just a factor that can be taken outside the integral? Or is everything inside the integral supposed to be forming the exponent?

 

[Ray Vickson]

Homework Statement

$$ \displaystyle \int_0^{\infty} e^{-x} \dfrac{a\sin ax - \cos ax}{1+a^2} da $$

Homework Equations

The Attempt at a Solution

Evaluating this using integration by parts will be a cumbersome process and I don't even think that would give me the answer. Substitutions aren't going to make this problem easier. What other methods should I apply?

I looked at your profile, and so cannot believe this is a homework problem you have been given in a course. For that reason, I am going to chance giving you a complete solution, a practice I normally avoid.

Let's change your unfortunate and rather horrible notation. What you call 'x' I will call 'b', and what you call 'a' I will call 'w'. So, you want to perform the integration
\int_0^{\infty} e^{-b} \frac{w \sin(b w) - \cos(bw)}{1+w^2} \, dw
First: the constant $e^{-b}$ plays no role, so omit it; you can just do the integral without it, then put it back again in the final answer.
Second: the integrand is an even function of $w$, so $\int_0^{\infty} = \frac{1}{2} \int_{-\infty}^{\infty}$, a fact that proves convenient later.

OK, so we want the integral
\int_{-\infty}^{\infty} \frac{ w \sin(bw) - \cos(bw)}{1+w^2} \, dw
WLOG, assume b > 0. We have $w \sin(bw) = w \text{Re}[-i e^{ibw}]$ and $-\cos(bw) = \text{Re}[-e^{ibw}]$, so the integrand is the real part of
F = \frac{-(1+iw) e^{ibw}}{1+w^2} = \frac{-(1+iw) e^{ibw}}{(1+iw)(1-iw)} =-i \frac{e^{ibw} }{w+i}.
Thus, it is enough to compute
J = \int_{-\infty}^{\infty} \frac{e^{ibw}}{w+i} \, dwThis is an improper integral, so let's replace it by
K_r = \int_{-\infty}^{\infty} e^{-r w^2} \frac{e^{ibw}}{w+i} \, dw
and then take the limit as $r \to 0+$.

The integrand of $K_r$ has a simple pole in the complex w-plane at the point $w = -i$. For $b > 0$ we may complete the contour in the upper half w-plane, to get
K_r = \lim_{R \to \infty} \int_{-R}^{R} e^{-r w^2} \frac{e^{ibw}}{w+i} \, dw<br /> = \lim_{R \to \infty} \oint_{\Gamma} e^{-r w^2} \frac{e^{ibw}}{w+i} \, dw,
where $\Gamma$ is the contour that goes along the real w-axis from $w = -R+i0$ to $w = +R+i0$, then goes counterclockwise from $R + i0$ to $-R + i0$ along the semicircle of radius $R$ in upper half of the complex $w$-plane. But the contour integral = 0 because the integrand is analytic inside $\Gamma$. Thus, $K_r = \lim_{R \to \infty} 0 = 0$, and this holds for all $r > 0$. Taking the limit we have
\int_{-\infty}^{\infty} \frac{e^{ibw}}{w+i} \, dw = \lim_{r \to 0} K_r = \lim_{r \to 0} 0 = 0.

Thus, we obtain
\int_{-\infty}^{\infty} \frac{ w \sin(bw) - \cos(bw)}{w^2+1} \, dw = 0,
provided that we regard it as the limit where $r \to 0+$.

 

[utkarshakash]

I looked at your profile, and so cannot believe this is a homework problem you have been given in a course. For that reason, I am going to chance giving you a complete solution, a practice I normally avoid.

Let's change your unfortunate and rather horrible notation. What you call 'x' I will call 'b', and what you call 'a' I will call 'w'. So, you want to perform the integration
\int_0^{\infty} e^{-b} \frac{w \sin(b w) - \cos(bw)}{1+w^2} \, dw
First: the constant $e^{-b}$ plays no role, so omit it; you can just do the integral without it, then put it back again in the final answer.
Second: the integrand is an even function of $w$, so $\int_0^{\infty} = \frac{1}{2} \int_{-\infty}^{\infty}$, a fact that proves convenient later.

OK, so we want the integral
\int_{-\infty}^{\infty} \frac{ w \sin(bw) - \cos(bw)}{1+w^2} \, dw
WLOG, assume b > 0. We have $w \sin(bw) = w \text{Re}[-i e^{ibw}]$ and $-\cos(bw) = \text{Re}[-e^{ibw}]$, so the integrand is the real part of
F = \frac{-(1+iw) e^{ibw}}{1+w^2} = \frac{-(1+iw) e^{ibw}}{(1+iw)(1-iw)} =-i \frac{e^{ibw} }{w+i}.
Thus, it is enough to compute
J = \int_{-\infty}^{\infty} \frac{e^{ibw}}{w+i} \, dwThis is an improper integral, so let's replace it by
K_r = \int_{-\infty}^{\infty} e^{-r w^2} \frac{e^{ibw}}{w+i} \, dw
and then take the limit as $r \to 0+$.

The integrand of $K_r$ has a simple pole in the complex w-plane at the point $w = -i$. For $b > 0$ we may complete the contour in the upper half w-plane, to get
K_r = \lim_{R \to \infty} \int_{-R}^{R} e^{-r w^2} \frac{e^{ibw}}{w+i} \, dw<br /> = \lim_{R \to \infty} \oint_{\Gamma} e^{-r w^2} \frac{e^{ibw}}{w+i} \, dw,
where $\Gamma$ is the contour that goes along the real w-axis from $w = -R+i0$ to $w = +R+i0$, then goes counterclockwise from $R + i0$ to $-R + i0$ along the semicircle of radius $R$ in upper half of the complex $w$-plane. But the contour integral = 0 because the integrand is analytic inside $\Gamma$. Thus, $K_r = \lim_{R \to \infty} 0 = 0$, and this holds for all $r > 0$. Taking the limit we have
\int_{-\infty}^{\infty} \frac{e^{ibw}}{w+i} \, dw = \lim_{r \to 0} K_r = \lim_{r \to 0} 0 = 0.

Thus, we obtain
\int_{-\infty}^{\infty} \frac{ w \sin(bw) - \cos(bw)}{w^2+1} \, dw = 0,
provided that we regard it as the limit where $r \to 0+$.

I appreciate your effort and patience for posting the complete solution. However, your method was too complex for me. I don't know anything about contours and related terms. The integral which I posted here is a part of another integral which forms the original problem and here's it.

Calculate \displaystyle \int_0^{\infty} e^{-x} \dfrac{\sin ax}{x} dx

Using Leibnitz's theorem for differentiation, I get
$$\displaystyle \int_0^{\infty} e^{-x} \cos ax dx $$
For evaluating this integral, I used the property(which is provided as a hint to the solution in my textbook) that
\displaystyle \int e^{ax} \cos bx dx = \dfrac{e^{ax}(a\cos bx + b \sin bx)}{a^2+b^2}

Substituting appropriate values of a and b in the given formula, I get the integral posted in the thread.

 

[vela]

You need to plug the limits in for $x$. That'll simplify the integrand to the point where it's easy to integrate with respect to $a$.

 

[Saitama]

Differentiating under the integral sign might help here, I think.

$$\displaystyle \frac{d}{dx}\int_0^{\infty} e^{-x} \dfrac{a\sin ax - \cos ax}{1+a^2} da = \int_0^{\infty} \frac{\partial}{\partial x} (e^{-x} \dfrac{a\sin ax - \cos ax}{1+a^2}) da = \int_0^{\infty}e^{-x}\cos ax da $$ which is non convergent.

So the original integral also does not exist.

I'm not 100% sure about this, though.

This reasoning is not correct. If you consider the integral $\displaystyle \int_0^{\infty} \frac{\sin (ax)}{x}\,dx$ and differentiate with respect to $a$, you get a non convergent integral but the original integral does exist!

BTW, @utkarshakash, how did you get to the integral in OP from the original integral?

 

[Curious3141]

This reasoning is not correct. If you consider the integral $\displaystyle \int_0^{\infty} \frac{\sin (ax)}{x}\,dx$ and differentiate with respect to $a$, you get a non convergent integral but the original integral does exist!

BTW, @utkarshakash, how did you get to the integral in OP from the original integral?

Thank you. Yes, that's why I stated I was unsure if it was correct in my first post, but I posted it in case someone else could use the idea to come up with a correct solution.

 

[haruspex]

Using Leibnitz's theorem for differentiation, I get
$$\displaystyle \int_0^{\infty} e^{-x} \cos ax dx $$
.

From there, can you not express the cos in complex exponentials and just integrate?

 

[utkarshakash]

@Pranav-Arora See post #8 of mine.

 

[vela]

From there, can you not express the cos in complex exponentials and just integrate?

Or integrate by parts. However, the OP doesn't even need to do that as the hint provided gives the result of the indefinite integration. The problem is that by not plugging in the limits for $x$, one is left with a complicated integrand in the last integration needed to get to the final solution.